A uniform 300-N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

Video transcript

Everyone in this problem. We have a 180 Newton hatch in the ceiling that is free to rotate about hinges on one of its ends. And we're asked to find what net upward force applied at the doors center okay. Is required to start opening the door. And were also asked what net forces exerted on the door by the hinges. Okay. All right. So let's draw a little diagram here. We're gonna take this to be the hinges. We're gonna draw the door just as a straight line. All right, now, which forces do we have? Well, the hatch has wait. Okay. And so we're gonna have the weight acting downwards. Now we have this upward force applied and it's applied at the door center, which is where we're going to have the weight. Okay, This is gonna be from the center of mass, which is the center of the door. So we have this upward force f you and then we also have net force exerted on the door by the hinges. And this is also going to be an upward force. Okay? We're gonna assume that this is an upward force in our picture and then when we do the calculation will see based on the sign that it may be the opposite way. Alright, so our coordinate system we're going to take up into the right is positive. We have our X and Y axis and we're going to consider the distance or the length of this hatch to be l. Okay? And then half the length will therefore be L over two. Okay. All right. So we're trying to find some information about forces. Okay, now let's recall this is an equilibrium problem. Okay. We have no net external forces acting on the system and so we know that the sum of the forces is going to be equal to zero. We also know that we have no net external torques. And so the sum of the torques is going to be equal to zero. Okay. Alright. So let's start with the sum of the forces. Now, we typically break this into the sum of the forces in the extraction some of the forces in the Y direction. In our case we only have forces in the Y direction. Ok, So those are the only ones we need to consider. So what are the forces in the Y direction? Well, in the positive Y direction we have this upward force. F. You? We have this force from the hinges. F. H. Okay. And then in the negative Y direction we have this weight. W. So we have minus W. Okay, And these are equal zero. Now, what we're trying to find here is we're trying to find this upward force F. U. And we're trying to find this force from the hinges. F. H. Okay, so in this equation we have two unknowns. We can't solve for two unknowns with one equation. So let's switch over to the torque equation and see what that gets us. Okay, now, we need to choose a pivot when we're looking at torque and we want to choose a pivot at one of these unknown forces. Okay? So that it doesn't contribute to the torque, it won't be in the equation. And then we're left with just um one variable to solve for in our equation. So let's take the pivot to be located at the Hinch. Okay? Alright. So if we take the pivot to be located at the hinge, what other torques do we have? Well we're gonna have the torque due to this upward force, we'll call it T you. Okay? And that's pointing upwards. So you can imagine it pushing kind of this hatch in the counterclockwise direction. So that's a positive torque. And then we have the torque from this weight w that's pointing downwards. You can imagine it pushing the hatch clockwise. Okay, so that's gonna be a negative torque. So we get minus the torque due to the weight. Okay? And this is gonna be equal to zero. Now the torque from this upward force, that's gonna be the distance from to the pivot. Okay. Which we've chosen to be at the hinge. So that's gonna be all over to times the force which is f. U. Times sine of the angle between the force And this hatch. Okay. Which is 90°. So that's our torque for the upward force. Now for the weight it's also going to be L over two away From the pivot. Okay, the forces W and it also makes a 90° angle with the hatch. Alright so we can divide both sides by L over two sine 90. And we're gonna be left with F. You -W. is equal to zero. Which tells us that the force, the upward force F. U. Is equal to W. Which is equal to 180 Newtons which were told in the problem. So now we have found our upward force. F. You. Okay? It's 180 newtons. So we've done part one of this problem outward force. Now we need to find the force exerted on the door by the hinges. Okay now we've said that that's F. H. And if we look back at our forces equation, okay now we know F. You, we know W. And so we can find F. H. And so F. H. Is going to be equal to w minus F. U. The weight. W. Is 100 and 80 newtons. Okay, again, we're told that in the problem. 180 newton. Hatch. F. U. Well, we found that this is also equal to 180 Nunes. And so the force exerted on the door by the Hinges is going to be zero Nunes. Mhm. Alright, So if we look at our answer choices, we see that the upward force we found was 180 newtons case, we're looking at one of these bottom three choices. We found that the force exerted on the door by the hinges was zero newtons. And so we have answer. E. Thanks everyone for watching. I hope this video helped see you in the next one.